Question

If \( x = \sqrt{2^{\text{cosec}^{-1} t}} \) and \( y = \sqrt{2^{\text{sec}^{-1} t}} (|t| \ge 1) \), then dy/dx is equal to :

• A. \( y/x \)
• B. \( -y/x \)
• C. \( -x/y \)
• D. \( x/y \)

Hint:

Whenever you see inverse trigonometric functions that sum to \( \pi/2 \) in exponents, try multiplying the functions to eliminate the parameter 't'. This simplifies the differentiation significantly. \

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question involves parametric equations with inverse trigonometric exponents. We need to find the derivative \( \frac{dy}{dx} \).

Step 2: Key Formula or Approach:
Recall the identity: \( \sec^{-1} t + \csc^{-1} t = \frac{\pi}{2} \).

Step 3: Detailed Explanation:
Square both given equations:
\( x^2 = 2^{\csc^{-1} t} \) and \( y^2 = 2^{\sec^{-1} t} \).
Multiply the two results: \[ x^2 y^2 = 2^{\csc^{-1} t} \cdot 2^{\sec^{-1} t} \] \[ x^2 y^2 = 2^{\csc^{-1} t + \sec^{-1} t} = 2^{\pi/2} \] Since \( 2^{\pi/2} \) is a constant, differentiate both sides with respect to \( x \): \[ \frac{d}{dx} (x^2 y^2) = 0 \] \[ 2x y^2 + x^2 \left( 2y \frac{dy}{dx} \right) = 0 \] \[ 2xy (y + x \frac{dy}{dx}) = 0 \] Since \( x, y \neq 0 \): \[ y + x \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x} \]

Step 4: Final Answer:
The derivative \( \frac{dy}{dx} \) is \( -\frac{y}{x} \).